adding one to infinity = infinity.
On 3/24/2024 1:13 PM, WM wrote:
Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
adding one to infinity = infinity.
If ℕ is complete no natural number can be added.
Sigh. You are misunderstanding infinity... I think wrt infinity, when
you hear the word "complete", your mind instantly thinks, "finite".
Infinity is not finite...
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
that both sets have the same number of elements.
Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...}
would double the number of its elements. Then there are more natural
numbers than were originally in ℕ.
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements. Then the completion of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements. Then there are more natural numbers than were originally in ℕ.
Regards, WM
WM was thinking very hard :
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn
be added?
Yes, it is the set of natural numbers.
Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
natürliche Zahlen als ℕ enthält.
I don't read German.
WM schrieb:
Sure
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
But you can do the following described by Cantor:
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
that both sets have the same number of elements.
Infinite sets don't have a "number of elements".
And indeed: there is a bijection from the set of natural numbers ℕ
to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
f: ℕ → 𝔼 , n ↦ 2n
This function is both injective (or one-to-one) and surjective (or
onto), thus it is bijective.
Then the completion of 𝔼 resulting in E = {1, 2, 3, 4, 5, 6, ...}
would double the number of its elements. Then there are more natural
numbers than were originally in ℕ.
Rubbish. The cardinality of an infinite set is described by an
transfinite cardinal number and not by a finite "number of elements".
Your problem is: You try to apply facts,
that hold for finite sets,
on infinite sets. That doesn't work.
On 3/24/2024 1:20 PM, WM wrote:
Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
On 3/24/2024 1:13 PM, WM wrote:
Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
adding one to infinity = infinity.
If ℕ is complete no natural number can be added.
Sigh. You are misunderstanding infinity... I think wrt infinity, when
you hear the word "complete", your mind instantly thinks, "finite".
Can you add a natural number to the set of all natura, numbers?
ℕ is not finite!
Infinity is not finite...
But logic has to be observed.
Right. So any natural number you give me, I can say that plus one.
Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :
WM schrieb:
Sure
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
Of course.
But you can do the following described by Cantor:
Irrelevant. Important is only this: You cannot add a natural number to
the set ℕ.
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
that both sets have the same number of elements.
Infinite sets don't have a "number of elements".
This cannot be denied: A bijection, if really existing, proves that one
of both sets has not one element more nor less than the other!
And indeed: there is a bijection from the set of natural numbers ℕ
to the set of even natural numbers 𝔼 = {2, 4, 6, ..}.
f: ℕ → 𝔼 , n ↦ 2n
This function is both injective (or one-to-one) and surjective (or
onto), thus it is bijective.
If so, that would result in: The set 𝔼 has not one element more nor less than the set ℕ.
Then the completion of 𝔼 resulting in E = {1, 2, 3, 4, 5, 6, ...}
would double the number of its elements. Then there are more natural
numbers than were originally in ℕ.
Rubbish. The cardinality of an infinite set is described by an
transfinite cardinal number and not by a finite "number of elements".
Here we do not use the rubbish of cardinality but the definition of
bijection proving that one of both sets has not one element more or less
than the other!
Your problem is: You try to apply facts,
I apply logic which is universally valid.
If the set 𝔼 = {2, 4, 6, ..} has not one element more or less than the
set ℕ = {1, 2, 3, ...}, then adding an element to 𝔼 destroys this state.
that hold for finite sets,
on infinite sets. That doesn't work.
Your problem is you deny logic which is universally valid.
Regards, WM
Le 24/03/2024 à 22:09, Richard Damon a écrit :
On 3/24/24 4:11 PM, WM wrote:
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove
that both sets have the same number of elements. Then the completion
of E resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number
of its elements. Then there are more natural numbers than were
originally in ℕ.
Yep, because "infinity" just doesn't obey the logic you are used to
Then do not talk about a bijection. A bijection between thwo sets proves
that one of both sets has not one element more nor less than the other!
Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)
This shows that card is in contradiction with basic logic.
despite that breaking most of you concepts of the logic of numbers.
It breaks the definition of bijection. Note: A bijection between thwo
sets proves that one of both sets has not one element more nor less than
the other!
Regards, WM
On 3/24/24 4:11 PM, WM wrote:
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that
both sets have the same number of elements. Then the completion of E
resulting in E = {1, 2, 3, 4, 5, 6, ...} would double the number of its
elements. Then there are more natural numbers than were originally in ℕ. >>
Yep, because "infinity" just doesn't obey the logic you are used to
Turns out that aleph0 = aleph0 - 1, = aleph0 + 1 = aleph0 / 2 =
aleph0 * 2 = aleph0 ^ 2 = square root (aleph0)
despite that breaking most of you concepts of the logic of numbers.
WM explained :
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be >> added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both
sets have the same number of elements.
Actually, same size set. "Number of elements" is better suited to
finite sets.
Then the completion of E resulting in
E = {1, 2, 3, 4, 5, 6, ...} would double the number of its elements.
That is not a completion of E. But still the same size set.
By your
sense of 'complete' the set of even numbers was already 'complete'
because no more even numbers could be 'added'.
Then there are more natural numbers than were originally in ℕ.
Nope.
Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :
[...]
Important is only this:
You cannot add a natural number to the set ℕ.
On 3/25/24 7:10 AM, WM wrote:
This cannot be denied: A bijection, if really existing, proves that one
of both sets has not one element more nor less than the other!
Nope,
just that they have the "same size", since for infinite sets, one
more or less is still the "same size", even twice the number of elements
can be the "same size", or the square of the number of elements is the
"same size"
That is just a property of infinite sets,
On 3/25/24 7:15 AM, WM wrote:
Then do not talk about a bijection. A bijection between thwo sets proves
that one of both sets has not one element more nor less than the other!
Nope, it proves they have the same size.
Nope, it shows that YOUR logic is not UNIVERSALLY VALID as you try to claim.
Nope, the definition of Bijection shows that the two sets have the same
size.
on 3/25/2024, WM supposed :
Insted of number of elements say: A bijection between thwo sets proves that >> one of both sets has not one element more nor less than the other!
No, it shows that two sets are the same size.
It has more elements than before, namely all odd numbers.
SETS DON'T CHANGE!!!
On 3/25/2024 7:10 AM, WM wrote:
Le 24/03/2024 à 22:02, Dieter Heidorn a écrit :
[...]
Important is only this:
You cannot add a natural number to the set ℕ.
not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n not.in ℕ
not.exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
⟦0,n⟧ not.fits ℕ
== not.exists 1.to.1.map ⟦0,n⟧ ⇉ ℕ
for each finiteⁿᵒᵗᐧᵂᴹ set S
exists finiteⁿᵒᵗᐧᵂᴹ ordinal n such that
⟦0,n⟧ not.fits S
== not.exists 1.to.1.map ⟦0,n⟧ ⇉ S
ℕ is not finiteⁿᵒᵗᐧᵂᴹ.
for each finiteⁿᵒᵗᐧᵂᴹ set S
not.exists 1.to.1.map S∪{Bob} > S
ℕ is not finiteⁿᵒᵗᐧᵂᴹ
for not.finiteⁿᵒᵗᐧᵂᴹ ℕ
exists 1.to.1.map ℕ∪{Bob} ⇉ ℕ
You can add Bob to ℕ
for not.finiteⁿᵒᵗᐧᵂᴹ ℕ
exists 1.to.1.map ℕ∪{Bob} ⇉ ℕ
You can add Bob to ℕ
Le 25/03/2024 à 20:38, Jim Burns a écrit :
[...]
You cannot add a natural number to ℕ.
But a bijection of ℕ with |E = {2, 4, 6, ...}
would prove that both sets have
the same number of elements.
Adding an element to |E destroys this state
and shows ℕ is larger than ℕ.
Contradiction!
On 3/25/2024 4:23 AM, WM wrote:
Le 24/03/2024 à 22:33, "Chris M. Thomasson" a écrit :
On 3/24/2024 1:20 PM, WM wrote:
Le 24/03/2024 à 21:16, "Chris M. Thomasson" a écrit :
On 3/24/2024 1:13 PM, WM wrote:
Le 24/03/2024 à 21:11, "Chris M. Thomasson" a écrit :
adding one to infinity = infinity.
If ℕ is complete no natural number can be added.
Sigh. You are misunderstanding infinity... I think wrt infinity,
when you hear the word "complete", your mind instantly thinks,
"finite".
Can you add a natural number to the set of all natura, numbers?
ℕ is not finite!
But it is complete. Sets are complete in ZF.
I think you are confusing complete with finite. This is not true of ℕ.
Infinity is not finite...
But logic has to be observed.
Right. So any natural number you give me, I can say that plus one.
And that is also already in ℕ.
This is why ℕ + 1 = ℕ
WM laid this down on his screen :
by showing injectivity and surjectivity. That means one-to-one
correspondence: one of both sets has not one element more nor less than the >> other!
For finite sets.
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
on 3/26/2024, WM supposed :
Le 25/03/2024 à 23:49, FromTheRafters a écrit :
WM laid this down on his screen :
by showing injectivity and surjectivity. That means one-to-one
correspondence: one of both sets has not one element more nor less than >>>> the other!
For finite sets.
For all bijections which deserve this name.
What name?
On 3/25/2024 5:04 PM, Jim Burns wrote:
On 3/25/2024 6:11 PM, WM wrote:
You cannot add a natural number to ℕ.
You cannot add a natural number to ℕ ==
ℕ holds all sizes of sets for which
changing by one element changes the set's size.
The set of odd numbers is infinite.
There are an infinite number of natural numbers.
infinity = infinity.
Density is another matter.
Density is another matter.
.01->.001 can represent infinity...
Its denser, so to speak.
Le 26/03/2024 à 01:04, Jim Burns a écrit :
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection
destroys this bijection.
WM used his keyboard to write :
Le 26/03/2024 à 14:39, FromTheRafters a écrit :
on 3/26/2024, WM supposed :
Le 25/03/2024 à 23:49, FromTheRafters a écrit :
WM laid this down on his screen :
by showing injectivity and surjectivity. That means one-to-one
correspondence: one of both sets has not one element more nor
less than the other!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Le 26/03/2024 à 01:04, Jim Burns a écrit :
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection.
Regards, WM
Phil Carmody submitted this idea :
FromTheRafters <[email protected]> writes:
WM used his keyboard to write :
Le 26/03/2024 à 14:39, FromTheRafters a écrit :
on 3/26/2024, WM supposed :
Le 25/03/2024 à 23:49, FromTheRafters a écrit :
WM laid this down on his screen :
by showing injectivity and surjectivity. That means one-to-one >>>>>>>> correspondence: one of both sets has not one element more nor
less than the other!
For finite sets.
For all bijections which deserve this name.
What name?
The name bijection.
All bijections deserve the name bijection.
Even the ones that are just bicuriousjections?
Yes. Math is an equivalence class opportunity employer. :D
ℕ and ℚ have the same infinity.
On 3/26/2024 9:24 AM, WM wrote:
Le 26/03/2024 à 01:04, Jim Burns a écrit :
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection
destroys this bijection.
Adding an element "destroys" the set.
Sets do not change.
There are other bijections for other sets.
And ℕ\{1,2} ⇉ ℕ\{1,2,3}
Et cetera, ad infinitum.
All bijections deserve the name bijection.
On 3/26/2024 6:11 AM, WM wrote:
Try to think better. Complete means complete, i.e., no element of ℕ is
available to be added to ℕ.
infinity + 1 = infinity
On 3/26/24 9:24 AM, WM wrote:
Le 26/03/2024 à 01:04, Jim Burns a écrit :But creates another that shows that they are still the same size.
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this bijection. >>
The existance of ANY bijection between the sets, proves them to be of
the same size.
Of course, you can't understand this,
Le 26/03/2024 à 16:40, Jim Burns a écrit :
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
Le 26/03/2024 à 17:04, Jim Burns a écrit :
On 3/26/2024 9:24 AM, WM wrote:
Le 26/03/2024 à 01:04, Jim Burns a écrit :
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
All + 1 is more than all.
There are not two
consecutive infinite sets in ℕ possible.
WM laid this down on his screen :
Le 26/03/2024 à 18:04, FromTheRafters a écrit :
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and O
deletes O.
Not showing a bijection is not the same as not having a bijection.
On 3/27/2024 12:10 PM, Jim Burns wrote:
On 3/27/2024 9:46 AM, WM wrote:
Le 26/03/2024 à 17:04, Jim Burns a écrit :
On 3/26/2024 9:24 AM, WM wrote:
Le 26/03/2024 à 01:04, Jim Burns a écrit :
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
All + 1 is more than all.
There are not two
consecutive infinite sets in ℕ possible.
Infiniteᵂᴹ is only
really.really.humongously finiteⁿᵒᵗᐧᵂᴹ
ℕ can't be any size such that
size+1 > size
because
N holds a finiteⁿᵒᵗᐧᵂᴹ ordinal larger than that size.
Therefore,
size(ℕ)+1 = size(ℕ)
ℕ + 1 = ℕ
So
adding one to the set of all natural numbers
means
that number was already there,
therefore it equals itself?
Le 26/03/2024 à 16:40, Jim Burns a écrit :
ℕ and ℚ have the same infinity.
Only if logic (every lossless exchange is lossless) is violated and
damaged, i.e., only in matheology.
Regards, WM
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
WM laid this down on his screen :
Le 26/03/2024 à 18:04, FromTheRafters a écrit :
All bijections deserve the name bijection.
Not those which violate logic in that lossless exchange between X and
O deletes O.
Not showing a bijection is not the same as not having a bijection.
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Between infinite sets there cannot exist any mapping because most
elements are dark. But we can assume that very simple mappings like f(x)
= x are true even for dark elements.
Therefore between the rational numbers and the natural numbers f(n) =
n/1 can be accepted, also f(n) = 1/n, but not f(n) = 2n.
Regards, WM
Le 27/03/2024 à 02:19, Richard Damon a écrit :
On 3/26/24 9:24 AM, WM wrote:
Le 26/03/2024 à 01:04, Jim Burns a écrit :But creates another that shows that they are still the same size.
ℕ and 𝔼 aren't sets for which
changing by one element changes the set's size.
Adding an element to one of the sets in bijection destroys this
bijection.
Only if the bijection was between potentially infinite sets.
The existance of ANY bijection between the sets, proves them to be of
the same size.
The existence of any injective but not surjective mapping shows that the
sets are not of same size. The existence of any surjective but not
injective mapping shows that the sets are not of same size.
Of course, you can't understand this,
I understand that your claims can only be satisfied by potential infinity.
Regards, WM
WM expressed precisely :
Showing a not bijection proves different sizes of sets.
No it doesn't.
On 3/27/24 5:50 PM, WM wrote:
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope. Cantor shows that there may well be "failed" attempts at a
bijection that actually don't tell us anything about the sizes of two infinite sets.
On 3/27/2024 6:50 AM, WM wrote:
Le 26/03/2024 à 19:56, "Chris M. Thomasson" a écrit :
On 3/26/2024 6:11 AM, WM wrote:But not the same set. What natnumber could be added to ℕ?
Try to think better. Complete means complete, i.e., no element of ℕ
is available to be added to ℕ.
infinity + 1 = infinity
Any natural can be added to ℕ, for ℕ + ℕ = ℕ,
On 3/27/24 9:57 AM, WM wrote:
Only if the bijection was between potentially infinite sets.
Which N and E are/*********************
I understand that your claims can only be satisfied by potential infinity.
We are talking about ACTUAL infinite sets, not just "potential",*******************************************************************
On 3/27/2024 9:38 AM, WM wrote:
Le 26/03/2024 à 16:40, Jim Burns a écrit :
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are the same size.
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
Le 28/03/2024 à 21:25, Jim Burns a écrit :
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No.
The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Under f(x) = 2x
we get the image {2, 4, 6, 8, 10, ..., 2ω].
The mapping restricted to the natural numbers shows
less evens than naturals.
But we can assume that
very simple mappings like f(x) = x are true
even for dark elements.
Therefore
between the rational numbers and the natural numbers
f(n) = n/1 can be accepted,
also f(n) = 1/n,
but not f(n) = 2n.
Where are darkᵂᴹ numbers introduced?
They are the places where
Bob rests when the mapping is finished.
On 3/28/2024 2:40 PM, WM wrote:
Le 28/03/2024 à 21:25, Jim Burns a écrit :
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Huh?
If it has a last element its NOT infinite...
Le 27/03/2024 à 18:54, Jim Burns a écrit :
On 3/27/2024 9:38 AM, WM wrote:
Le 26/03/2024 à 16:40, Jim Burns a écrit :
ℕ and ℚ have the same infinity.
Only if
logic (every lossless exchange is lossless)
is violated and damaged, i.e.,
only in matheology.
ℕ and ℚᶠʳᵃᶜ are the same size.
ℕ and ℕ are the same size.
ℚ and ℚ are the same size.
Removing a proper fraction decreases ℚ but leaves it larger than ℕ. When the size changes it cannot remain the same.
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Between infinite sets there cannot exist any mapping because most
elements are dark. But we can assume that very simple mappings like f(x)
= x are true even for dark elements.
Therefore between the rational numbers and the natural numbers f(n) =
n/1 can be accepted, also f(n) = 1/n, but not f(n) = 2n.
Regards, WM
Regards, WM
Le 28/03/2024 à 01:57, Richard Damon a écrit :
On 3/27/24 9:57 AM, WM wrote:
*********************Only if the bijection was between potentially infinite sets.
Which N and E are/*********************
I understand that your claims can only be satisfied by potential
infinity.
*******************************************************************
We are talking about ACTUAL infinite sets, not just "potential",*******************************************************************
Regards, WM
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
The mapping restricted to the natural numbers shows less evens than naturals.
On 3/28/2024 2:40 PM, WM wrote:
Le 28/03/2024 à 21:25, Jim Burns a écrit :
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
Huh? If it has a last element its NOT infinite...
Le 27/03/2024 à 23:37, FromTheRafters a écrit :
WM expressed precisely :
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Regards, WM
The clown WM drivels:
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Le 28/03/2024 à 01:59, Richard Damon a écrit :
On 3/27/24 5:50 PM, WM wrote:
Showing a not bijection proves different sizes of sets.
Why is that more meaningful than Cantor's bijections?
Nope. Cantor shows that there may well be "failed" attempts at a
bijection that actually don't tell us anything about the sizes of two
infinite sets.
Why do you think so?
Regards, WM
On 3/29/2024 7:13 AM, Tom Bola wrote:
Chris M. Thomasson schrieb:
On 3/28/2024 2:40 PM, WM wrote:
Le 28/03/2024 à 21:25, Jim Burns a écrit :
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No. The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a >>>> last element.
Huh? If it has a last element its NOT infinite...
Oops, you might also write that this way: { ω, 1, 2, 3, 4, 5, ... }.
I got confused. I thought that ω was WM's largest natural number, its
last element, so to speak.
On 3/29/2024 4:29 PM, Tom Bola wrote:
Chris M. Thomasson schrieb:
On 3/29/2024 7:13 AM, Tom Bola wrote:
Chris M. Thomasson schrieb:
On 3/28/2024 2:40 PM, WM wrote:
Le 28/03/2024 à 21:25, Jim Burns a écrit :
On 3/27/2024 5:50 PM, WM wrote:
Le 27/03/2024 à 15:23, FromTheRafters a écrit :
Not showing a bijection is not the same as
not having a bijection.
Showing a not bijection proves
different sizes of sets.
I'll call that a Mückenheim.set.
Mückenheim.sets are finiteⁿᵒᵗᐧᵂᴹ sets.
No.
The set {1, 2, 3, 4, 5, ..., ω} is infinite,
although there is a last element.
Huh? If it has a last element its NOT infinite...
I know for sure that WM thinks
there is a largest natural number.
That must be his last element.
Still, I do not know why he thinks that way.
Oh well. Shit happens.
And, it can be shown that ℕ and ℚ are the same size.
Nope, Removing a single element from an infinite set doesn't change its
size.
On 3/28/24 3:55 PM, WM wrote:
Le 27/03/2024 à 23:37, FromTheRafters a écrit :
WM expressed precisely :
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Because it is showable that you CAN make non-working attempts at
bijections on infinite sets that you can also show working bijections,
I know for sure that WM thinks there is a largest natural number. That
must be his last element. Still, I do not know why he thinks that way.
I think that, in WM's telling,
ω is _immediately after_ the largest natural number,
I think that WM thinks that, in our telling,
ω is a natural number
Why he thinks that
is a question for psychologists.
Tom Bola a écrit :
The clown WM drivels:No. Cantor accepts ω, ω + 2, ω + 4, ... These numbers can be realized
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of
ordinals,
i.e. natural numbers and also a limit ordinal like ω.
The mapping restricted to the natural numbers shows less evens than naturals.
this set would contain
the union of W1 = {2, 4, 6, 8, 10, ...} of card ω and W2 = {2ω} of card 1. >>
by repeated addition. Mutiplication is a shorthand of addition.
Prof. Dr. Mückenheim taught:
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of ordinals,
i.e. natural numbers and also a limit ordinal like ω.
The mapping restricted to the natural numbers shows less evens than naturals.
this set would contain
the union of W1 = {2, 4, 6, 8, 10, ...} of card ω and W2 = {2ω} of card 1.
Le 29/03/2024 à 18:05, Richard Damon a écrit :
On 3/28/24 3:55 PM, WM wrote:
Le 27/03/2024 à 23:37, FromTheRafters a écrit :
WM expressed precisely :
Showing a not bijection proves different sizes of sets.
No it doesn't.
Why not?
Because it is showable that you CAN make non-working attempts at
bijections on infinite sets that you can also show working bijections,
Only if the Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
could disappear by exchange with Xs which is ipossible.
Regards, WM
Le 29/03/2024 à 14:29, Richard Damon a écrit :
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand logic.
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
Nope, Removing a single element from an infinite set doesn't change
its size.
That is believed by some stupids who are too dense to understand logic. Removing one element makes the set having one element less than before.
Regrads, WM
Le 30/03/2024 à 13:56, Jim Burns a écrit :
I think that, in WM's telling,
ω is _immediately after_ the largest natural number,
like the smallest unit fraction is existing.
I think that WM thinks that, in our telling,
ω is a natural number
No.
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can
be added?
If so, then the bijection of ℕ with E = {2, 4, 6, ...} would prove that both sets have the same number of elements.
Then the completion of E resulting in E* = {1, 2, 3, 4, 5, 6, ...}[.] [Hence the number of E*'s elements] would [be twice] the number of [E's] elements.
Then there are more natural numbers than were originally in ℕ.
Piss off already, blithering retard...
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
On 3/30/24 12:01 PM, WM wrote:
Le 29/03/2024 à 14:29, Richard Damon a écrit :
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand logic.
The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
So?
That isn't the Bijection we are looking for.
Le 30/03/2024 à 20:04, Jim Burns a écrit :
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
On 03/31/2024 06:19 AM, Tom Bola wrote:
WM schrieb:
Le 30/03/2024 à 20:04, Jim Burns a écrit :
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
This is a sequence of all ordinals including limit ordinals, while your
attempt wants to make statements about the natural numbers 1, 2, 3, ...
Just think of it as a model of modularity,
that modules are of equal size, that modules
have a grain or equi-partitioning, as the
number or count of those increases, and
then as in the infinite limit and continuum limit,
that the boundaries of the modules maintain their
proportion, of course though this is still very "natural".
The modularity of numbers is a key term with trichotomy.
Otherwise this sort of f(n) = n, f(n) = 2n, and so on,
simply reflect as a function and functional relation,
how it so happens that systems of equations include
the algebra of functions, keeping things simple.
Whether natural numbers include infinite numbers,
the infinitely-grand, has that according to naive
expansion of comprehension they do. It's just the
regular ordinary opinion of regular ordinary set
theory, they don't.
There are others, ....
Es gibt anderen, ....
"Sein ein Hengst, nicht ein Zwerg."
On 03/31/2024 06:19 AM, Tom Bola wrote:
WM schrieb:
Le 30/03/2024 à 20:04, Jim Burns a écrit :
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
This is a sequence of all ordinals including limit ordinals, while your
attempt wants to make statements about the natural numbers 1, 2, 3, ...
Just think of it as a model of modularity,
that modules are of equal size, that modules
have a grain or equi-partitioning, as the
number or count of those increases, and
then as in the infinite limit and continuum limit,
that the boundaries of the modules maintain their
proportion, of course though this is still very "natural".
The modularity of numbers is a key term with trichotomy.
Otherwise this sort of f(n) = n, f(n) = 2n, and so on,
simply reflect as a function and functional relation,
how it so happens that systems of equations include
the algebra of functions, keeping things simple.
Whether natural numbers include infinite numbers,
the infinitely-grand, has that according to naive
expansion of comprehension they do. It's just the
regular ordinary opinion of regular ordinary set
theory, they don't.
There are others, ....
Es gibt anderen, ....
"Sein ein Hengst, nicht ein Zwerg."
If there are all points on the real line permanently existing, then
there is a point next to zero.
Am 29.03.2024 um 17:31 schrieb Tom Bola:
Sets are not ordered, so <bla>
Check: https://en.wikipedia.org/wiki/Partially_ordered_set
and: https://mathworld.wolfram.com/OrderedSet.html
Sets are not ordered, so <bla>
Am 29.03.2024 um 17:31 schrieb Tom Bola:
Sets are not ordered, so <bla>
Check: https://en.wikipedia.org/wiki/Partially_ordered_set
and: https://mathworld.wolfram.com/OrderedSet.html
Le 30/03/2024 à 20:04, Jim Burns a écrit :
Hold back just a moment on assigning labels.
Is this a fair description of
what you think we think?
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
Le 30/03/2024 à 19:47, Richard Damon a écrit :
On 3/30/24 12:01 PM, WM wrote:
Le 29/03/2024 à 14:29, Richard Damon a écrit :
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand
logic. The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
So?
That isn't the Bijection we are looking for.
It is true for every mapping.
Regards, WM
On 3/31/2024 1:38 PM, Jim Burns wrote:
On 3/31/2024 8:20 AM, WM wrote:
[...][...]
⟦0,φ⦆ = {λ ordinal: 0 ≤ λ < φ }
φ ∈ ⟦0,φ⦆ ∌ φ+1
On 3/31/2024 8:20 AM, WM wrote:
[...][...]
No totally.ordered ⟦0,φ⦆ ⟦0,φ+1⦆ exist such that
every non-empty subset of ⟦0,φ⦆ has both
a least and a greatest element in the subset
and
NOT every non-empty subset of ⟦0,φ+1⦆ has both
a least and a greatest element in the subset.
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
I read your response as a confirmation of
what I think what you think we think.
We (not.WM) do not think that.
Above, you can see reasons we do not think that.
"Infinite" is different.
"Infinite" is not simply "humongous" in dress.up.
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 = aleph_0
+ aleph_0 = aleph_0.
On 3/31/2024 8:20 AM, WM wrote:
ψ not.exists such that ψ+1 = ω
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
I read your response as a confirmation of
what I think what you think we think.
Am 30.03.2024 um 17:09 schrieb WM:
If there are all points on the real line permanently existing, then
there is a point next to zero.
No, there is no point "next to zero".
If x is a point "close to zero", then the point x/2 is even closer to
zero,
On 3/31/24 8:15 AM, WM wrote:
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand
logic. The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
So?
That isn't the Bijection we are looking for.
It is true for every mapping.
Nope, only if you try to biject WITH a set, and not between two
DIFFERENT sets.
Le 31/03/2024 à 17:38, Jim Burns a écrit :
On 3/31/2024 8:20 AM, WM wrote:If all [ordinal] numbers exist, then also this number exists.
[No ordinal} ψ exists such that ψ+1 = ω
What else should happen by subtracting 1 from ω?
Le 31/03/2024 à 17:38, Jim Burns a écrit :
On 3/31/2024 8:20 AM, WM wrote:If all [ordinal] numbers exist, then also this number exists.
[No ordinal} ψ exists such that ψ+1 = ω
What else should happen by subtracting 1 from ω?
Le 30/03/2024 à 23:53, Moebius a écrit :
That
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 = aleph_0 + aleph_0 = aleph_0.
Le 31/03/2024 à 17:14, Moebius a écrit :
Am 30.03.2024 um 17:09 schrieb WM:If there <bla>
If there are all points on the real line permanently existing, then
there is a point next to zero.
No, there is no point "next to zero".
If x is a point "close to zero", then the point x/2 is even closer to zero, >>Not for all dark points x does x/2 exist.
Am 01.04.2024 um 17:04 schrieb WM:
Le 30/03/2024 à 23:53, Moebius a écrit :
That
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 = aleph_0 + aleph_0 = aleph_0.
are trivial facts in set theory.
See: https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_addition
and: https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_multiplication
Hope this helps.
Le 31/03/2024 à 17:38, Jim Burns a écrit :
On 3/31/2024 8:20 AM, WM wrote:
Le 30/03/2024 à 20:04, Jim Burns a écrit :
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
I read your response as a confirmation of
what I think what you think we think.
Cantor thinks so.
ψ not.exists such that ψ+1 = ω
If all numbers exist,
then also this number exists.
What else should happen by subtracting 1 from ω?
Moebius schrieb:
Am 01.04.2024 um 17:04 schrieb WM:But WM has found all that to be wrong (by the laws of "natural logic") and feels to be on earth to help all ignorant mankind...
Le 30/03/2024 à 23:53, Moebius a écrit :
That
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 = aleph_0 + aleph_0 = aleph_0.
are trivial facts in set theory.
See: https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_addition
and: https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_multiplication >>
Hope this helps.
Und vielleicht NOCH etwas deutlicher:
| Ax e IR: Ey e IR: y = x/2 && y < x.
==========================================================
Geometrsch stellt sich das so dar, Mückenhirn: Jede noch so kleine
Strecke kann (mittig) geteilt werden. "Dunkle (unteilbare) Strecken"
gibt es (in der Geometrie) nicht.
Le 31/03/2024 à 20:18, Richard Damon a écrit :
On 3/31/24 8:15 AM, WM wrote:
And, it can be shown that ℕ and ℚ are the same size.
That is believed by some stupids who are too dense to understand
logic. The Os in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
..
cannot disappear by exchange with Xs.
So?
That isn't the Bijection we are looking for.
It is true for every mapping.
Nope, only if you try to biject WITH a set, and not between two
DIFFERENT sets.
The different sets are ℕ and ℚ. The bijection with the first column does not change that.
Regards, WM
On 4/1/2024 11:34 AM, WM wrote:
ψ not.exists such that ψ+1 = ω
If all numbers exist,
then also this number exists.
All ordinals which exist exist.
All ordinals which not.exist not.exist.
Why do you (WM) think
ψ: ψ+1 = ω is one of the former, not the latter?
On 4/1/24 11:37 AM, WM wrote:
The different sets are ℕ and ℚ. The bijection with the first column does >> not change that.
Yes it does, as you are not "moving" the O out of the set of Q.
That makes a difference.
Le 01/04/2024 à 18:34, Jim Burns a écrit :
On 4/1/2024 11:34 AM, WM wrote:
I think that WM thinks that, in our telling,
one gets to ω
by going out out out to "infinity"
1, 2, 3, ... ω, ω + 1, ... [Cantor, 1882]
I read your response as a confirmation of
what I think what you think we think.
Cantor thinks so.
You (WM) think Cantor thinks so.
Also,
you call _us_ "matheologians" and "stupid",
indicating that
you think we think what you think Cantor thinks.
We do not
think what you think Cantor thinks.
ψ not.exists such that ψ+1 = ω
If all numbers exist,
then also this number exists.
All ordinals which exist exist.
All ordinals which not.exist not.exist.
Why do you (WM) think
ψ: ψ+1 = ω is one of the former, not the latter?
Because something must be there
on the ordinal axis.
All ordinals which exist exist.
All ordinals which not.exist not.exist.
Why do you (WM) think
ψ: ψ+1 = ω is one of the former, not the latter?
Because something must be there
on the ordinal axis.
Le 02/04/2024 à 01:03, Richard Damon a écrit :
On 4/1/24 11:37 AM, WM wrote:
The different sets are ℕ and ℚ.
The bijection with the first column
does not change that.
Yes it does,
as you are not "moving" the O out of the set of Q.
That makes a difference.
It does not make a difference.
It only shows that there is no bijectio.
ℚᶠʳᵃᶜ
Am 02.04.2024 um 19:51 schrieb Jim Burns:
ℚᶠʳᵃᶜ
Very nice idea!
On 4/2/2024 4:52 PM, Moebius wrote:
Am 02.04.2024 um 19:51 schrieb Jim Burns:
Very nice idea!
ℚᶠʳᵃᶜ
Thank you.
Of course, "i/j as fraction not rational"
is WM's idea.
On 04/02/2024 02:40 PM, Jim Burns wrote:
On 4/2/2024 4:52 PM, Moebius wrote:
Am 02.04.2024 um 19:51 schrieb Jim Burns:
ℚᶠʳᵃᶜ
Very nice idea!
Thank you.
Of course, "i/j as fraction not rational"
is WM's idea. I'm only trying to keep
the expression of those ideas from making
the ideas themselves hard to discern.
My exploration of Unicode is a little treat
I give myself for having to repeat the same thing
over and over and over.
Doesn't it repeat itself,
and we just echo it?
On 04/02/2024 04:25 PM, Jim Burns wrote:
On 4/2/2024 6:06 PM, Ross Finlayson wrote:
On 04/02/2024 02:40 PM, Jim Burns wrote:
My exploration of Unicode is a little treat
I give myself for having to repeat the same thing
over and over and over.
Doesn't it repeat itself,
and we just echo it?
Speech with no speaker?
Not a thing I've seen around here, at least.
Perhaps that is the target which ChatGPT and such as
are aiming at. I ask you to take me at my word
when I claim to be a Natural.Intelligence,
something between a chimp and an angel.
Yeah, we know you're sort of genius.
Yet, are you a platonist?
A mathematical platonist?
On 4/2/2024 3:36 AM, WM wrote:
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
Because something must be there
on the ordinal axis.
What we think is that
ω is the first ordinal with _infinite_ ⟦0,ω⟧
What we think is that
each non.empty subset S of _finite_ ⟦0,ψ⟧
has both a least and a greatest element in S
Lemma.
If ⟦0,ψ⟧ is finite,
then ⟦0,ψ+1⟧ = ⟦0,ψ⟧∪{ψ+1} is finite.
For each ordinal ψ visibleᵂᴹ or darkᵂᴹ
⟦0,ψ+1⟧ ≠ ⟦0,ω⟧
Le 02/04/2024 à 15:35, Jim Burns a écrit :
What we think is that
each non.empty subset S of _finite_ ⟦0,ψ⟧
has both a least and a greatest element in S
Lemma.
If ⟦0,ψ⟧ is finite,
then ⟦0,ψ+1⟧ = ⟦0,ψ⟧∪{ψ+1} is finite.
That is true for potential infinity.
In actual infinity however there is
no unoccupied point below ω.
For each ordinal ψ visibleᵂᴹ or darkᵂᴹ
⟦0,ψ+1⟧ ≠ ⟦0,ω⟧
Then not all numbers smaller than ω do exist,
contrary to set theory.
Le 02/04/2024 à 17:51, Jim Burns a écrit :
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why
first bijecting n and n/1 should destroy
an existing bijection?
WM presented the following explanation :
Le 02/04/2024 à 17:51, Jim Burns a écrit :
On 4/2/2024 3:36 AM, WM wrote:
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an existing
bijection?
Your 'trick' only fails to demonstrate a bijection. Failing to
demonstrate a bijection does not mean that there is no bijection, only
that your 'trick' doesn't work to that end.
On 4/3/2024 9:32 AM, WM wrote:
Le 02/04/2024 à 17:51, Jim Burns a écrit :
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
See below,
for each k ∈ ℕ its own iₖ/jₖ ∈ ℚᶠʳᵃᶜ
for each i/j ∈ ℚᶠʳᵃᶜ its own kᵢⱼ ∈ ℕ
Your trick gives incorrect results.
Bijecting n and n/1 do not "destroy"
bijections between ℕ and ℚᶠʳᵃᶜ
Le 03/04/2024 à 15:59, FromTheRafters a écrit :
WM presented the following explanation :
Le 02/04/2024 à 17:51, Jim Burns a écrit :
On 4/2/2024 3:36 AM, WM wrote:
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an
existing bijection?
Your 'trick' only fails to demonstrate a bijection. Failing to
demonstrate a bijection does not mean that there is no bijection, only
that your 'trick' doesn't work to that end.
Explain why first bijecting n and n/1 should destroy an existing bijection!
Regards, WM
On 4/4/24 5:33 AM, WM wrote:
Le 03/04/2024 à 15:59, FromTheRafters a écrit :It doesn't, Bijections are always between two DISTINCT sets, not a set
WM presented the following explanation :
Le 02/04/2024 à 17:51, Jim Burns a écrit :
On 4/2/2024 3:36 AM, WM wrote:
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an
existing bijection?
Your 'trick' only fails to demonstrate a bijection. Failing to
demonstrate a bijection does not mean that there is no bijection, only
that your 'trick' doesn't work to that end.
Explain why first bijecting n and n/1 should destroy an existing bijection! >>
and a piece of itself thought of as a set.
Le 04/04/2024 à 14:19, Richard Damon a écrit :
On 4/4/24 5:33 AM, WM wrote:"In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is
Le 03/04/2024 à 15:59, FromTheRafters a écrit :It doesn't, Bijections are always between two DISTINCT sets, not a set
WM presented the following explanation :
Le 02/04/2024 à 17:51, Jim Burns a écrit :
On 4/2/2024 3:36 AM, WM wrote:
If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy
an existing bijection?
Your 'trick' only fails to demonstrate a bijection. Failing to
demonstrate a bijection does not mean that there is no bijection,
only that your 'trick' doesn't work to that end.
Explain why first bijecting n and n/1 should destroy an existing
bijection!
and a piece of itself thought of as a set.
equinumerous to A. Explicitly, this means that there exists a bijective function from A onto some proper subset B of A." Wikipedia.
Regards, WM
Bijections are always between two DISTINCT sets,
not a set and a piece of itself thought of as a set.
Le 03/04/2024 à 23:48, Jim Burns a écrit :
Bijecting n and n/1 do not "destroy"
bijections between ℕ and ℚᶠʳᵃᶜ
Mapping n/1 in the fractions
shows that no bijection is possible.
On 04/03/2024 06:13 AM, Jim Burns wrote:
[...]
Iota-values:
the word "iota" means "smallest non-zero value".
Iota-values:
the word "iota" means "smallest non-zero value".
Real-values:
all the values between negative infinity and infinity.
On 4/4/24 9:07 AM, WM wrote:
It doesn't, Bijections are always between two DISTINCT sets, not a set"In mathematics, a set A is Dedekind-infinite (named after the German
and a piece of itself thought of as a set.
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. Explicitly, this means that there exists a bijective
function from A onto some proper subset B of A." Wikipedia.
Right, but that "Proper Subset" is considered as an independent item,
not as just pieces of the original set.
always ⟦0,m+1⟧ ≠ ⟦0,ω⟧
Le 04/04/2024 à 15:22, Richard Damon a écrit :
On 4/4/24 9:07 AM, WM wrote:
It doesn't, Bijections are always between two DISTINCT sets, not a"In mathematics, a set A is Dedekind-infinite (named after the German
set and a piece of itself thought of as a set.
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. Explicitly, this means that there exists a
bijective function from A onto some proper subset B of A." Wikipedia.
Right, but that "Proper Subset" is considered as an independent item,
not as just pieces of the original set.
Nevertheless it is a piece of the original set.
Regards, WM
On 04/04/2024 12:01 PM, Jim Burns wrote:
On 4/3/2024 10:13 PM, Ross Finlayson wrote:
Iota-values:
the word "iota" means "smallest non-zero value".
That which you use iota to describe
is not the continuum.
Real-values:
all the values between negative infinity and infinity.
There are several ambiguities in that description.
How about instead
Real.values:
least.upper.bounds of
bounded.non.empty.sets of
differences.of.ratios of
ordinals not.fitting.predecessors.
It's simple that the continuum limit, a limit of functions,
as we used to say modeling a function as a limit of a family
of functions, like for Dirac delta, these days it's often
called a "generalized distribution", such a function, with
its real analytical character, here the Equivalency Function
is unlike the Dirac delta in that it's not just an infinite
spike at the origin only under which is area one, while,
it is integrable, which is particularly unique for a function
from a discrete domain, and under it is area one, thus that
it's a generalized distribution if you will, while also it's
a continuum limit with the usual meaning of the words.
Then, that its range has extent, density, completeness, measure,
particularly completeness and measure, in [0,1], establishes
its range is a continuous domain, that instead of one or
the other of line-reals or field-reals, there are both.
Le 04/04/2024 à 17:03, Jim Burns a écrit :
always ⟦0,m+1⟧ ≠ ⟦0,ω⟧
The difference between ⟦0,m+1⟧ and ⟦0,ω⟧ is
how large?
Is it ω for every m?
Then what are the ordinals between m and ω?
They are dark.
On the other hand
no ordinal fits between ℕ and ω.
Dark ordinals reach till ω.
Agreed?
The clown WM drivels:
The set {1, 2, 3, 4, 5, ..., ω} is infinite, although there is a
last element.
One may construct such a set "N-with-omega".
Under f(x) = 2x we get the image {2, 4, 6, 8, 10, ..., 2ω}.
One may also define such a function, which domain contains both sort of ordinals,
i.e. natural numbers and also a limit ordinal like ω.
The mapping restricted to the natural numbers shows less evens than naturals.
The above sentence is very idiotic nonsense, [...]
You are way too dense for even simplest thinking and math...
WM explained on 4/4/2024 :
Explain why first bijecting n and n/1 should destroy an existing bijection!
You still seem to think that sets change. If you mean 'n' is an element
of the naturals then of course N bijects with the naturals as embedded
in Q.
Also, the complement of the naturals over one in Q is the same
size as the proper subset you created.
Le 05/04/2024 à 12:57, FromTheRafters a écrit :
WM explained on 4/4/2024 :
Explain why first bijecting n and n/1 should destroy an existing
bijection!
You still seem to think that sets change. If you mean 'n' is an
element of the naturals then of course N bijects with the naturals as
embedded in Q.
Of course. But if someone doubts it, I could directly map the naturals
n/1 to the fractions with the result that there is no bijection.
Also, the complement of the naturals over one in Q is the same size as
the proper subset you created.
No, that is disproved by the remaining Os.
Regards, WM
On 04/05/2024 11:04 AM, Jim Burns wrote:
[...]
I usually write that n/d with
n as "numerator" and d as "denominator".
It's not identified which f(n) is 1/root2,
only that it exists.
There's that for each r in [0,1],
exists n s.t. f(n) = r,
and f^-1(r) = n, mostly infinite.
There's not much said except that
d goes to infinity, and n goes to d.
(Thus that it's not just zero.)
Hint@Mückenheim: {2, 4, 6, 8, 10, ..., 2ω} = {2n : n e IN} u {2ω}.
On 4/5/2024 5:06 AM, WM wrote:
The difference between ⟦0,m+1⟧ and ⟦0,ω⟧ is
how large?
Is it ω for every m?
∀k,m ∈ ⟦0,ω⦆: k+m ∈ ⟦0,ω⦆
So, yes.
no ordinal fits between ℕ and ω.
ℕ = ⟦0,ω⦆
So, we agree on something.
Dark ordinals reach till ω.
Agreed?
⟦0,ξ⟧ which reaches 'til ω both
is a Mückenheim.set and is not a Mückenheim.set.
Agreed?
On 4/6/24 9:26 AM, WM wrote:
Le 05/04/2024 à 12:57, FromTheRafters a écrit :
WM explained on 4/4/2024 :
Explain why first bijecting n and n/1 should destroy an existing
bijection!
You still seem to think that sets change. If you mean 'n' is an
element of the naturals then of course N bijects with the naturals as
embedded in Q.
Of course. But if someone doubts it, I could directly map the naturals
n/1 to the fractions with the result that there is no bijection.
No, not "No Bijection", but that mapping isn't a bijection.
No, that is disproved by the remaining Os.
Which only shows that this one mapping doesn't work.
And, when you try it within one set, as opposed to between two sets,
Le 06/04/2024 à 15:40, Richard Damon a écrit :
On 4/6/24 9:26 AM, WM wrote:
Le 05/04/2024 à 12:57, FromTheRafters a écrit :
WM explained on 4/4/2024 :
Explain why first bijecting n and n/1 should destroy an existing
bijection!
You still seem to think that sets change. If you mean 'n' is an
element of the naturals then of course N bijects with the naturals
as embedded in Q.
Of course. But if someone doubts it, I could directly map the
naturals n/1 to the fractions with the result that there is no
bijection.
No, not "No Bijection", but that mapping isn't a bijection.
That mapping is Cantor's proposal. But for every other mapping, the O's
would also remain. All O's! It is th lossless exchange which proves it.
It is Cantor's famaous mapping, more than a century believed to be a bijection.
No, that is disproved by the remaining Os.
Which only shows that this one mapping doesn't work.
And, when you try it within one set, as opposed to between two sets,
If it operates, it must operate within one set too.
Regards, WM
Le 05/04/2024 à 19:03, Jim Burns a écrit :
On 4/5/2024 5:06 AM, WM wrote:
Dark ordinals reach till ω.
Agreed?
⟦0,ξ⟧ which reaches 'til ω both
is a Mückenheim.set and is not a Mückenheim.set.
Agreed?
There are no Mückenheim sets.
But we can use the ordinal axis
as Cantor has described it
0, 1, 2, 3, ...,
ω, ω + 1, ..., ω + k, ...,
ω + ω (= ω2), ω2 + 1, ..
and multiply 0, 1, 2, 3, ..., ω, by 2.
The difference between ⟦0,m+1⟧ and ⟦0,ω⟧ is
how large?
Is it ω for every m?
∀k,m ∈ ⟦0,ω⦆: k+m ∈ ⟦0,ω⦆
So, yes.
Then it is variable, not a fixed number.
Actually infinite sets are constant.
Potentially infinite collections are variable.
On 4/6/2024 9:44 AM, WM wrote:
But we can use the ordinal axis
as Cantor has described it
0, 1, 2, 3, ...,
ω, ω + 1, ..., ω + k, ...,
ω + ω (= ω2), ω2 + 1, ..
and multiply 0, 1, 2, 3, ..., ω, by 2.
Two times anything on the first row
is something on the first row.
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping, the O's
would also remain. All O's! It is th lossless exchange which proves it.
Cantor's proposal is between members of two distinct sets.
It is Cantor's famaous mapping, more than a century believed to be a
bijection.
But HIS does work, when you do it right.
If it operates, it must operate within one set too.
Why?
Le 06/04/2024 à 15:58, Richard Damon a écrit :
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping, the
O's would also remain. All O's! It is th lossless exchange which
proves it.
Cantor's proposal is between members of two distinct sets.
No. He does not specify that. And there is no reason to do so, except
that it can be used to contradict the ridiculous nonsense that there are
as many fractions as prime numbers.y
+
It is Cantor's famaous mapping, more than a century believed to be a
bijection.
But HIS does work, when you do it right.
No, his bijection works only for potential infinity applying the "...".
I show that his mapping does not work for the complete actually infinite sets. He uses "and so on". Why does any intelligent mind believe that? I
show that the remainder will never decrease. There is no belief
required. It is provable fact.
If it operates, it must operate within one set too.
Why?
Because there are as many naturals in ℕ as in ℚ. Precisely as many. But only my approach shows that they are less than the fractions.
Regards, WM
On 4/6/24 3:40 PM, WM wrote:
Le 06/04/2024 à 15:58, Richard Damon a écrit :
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping, the
O's would also remain. All O's! It is th lossless exchange which
proves it.
Cantor's proposal is between members of two distinct sets.
No. He does not specify that. And there is no reason to do so, except
that it can be used to contradict the ridiculous nonsense that there are
as many fractions as prime numbers.y
But he DOES, as he talks about the two SETS of numbers that are matched up.
And yes, the size of the set of all fractions is EXACTLY of the same
size as the set of all Prime Numbers, and that size is Aleph_0.
Le 06/04/2024 à 22:03, Richard Damon a écrit :
On 4/6/24 3:40 PM, WM wrote:
Le 06/04/2024 à 15:58, Richard Damon a écrit :
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping, the
O's would also remain. All O's! It is th lossless exchange which
proves it.
Cantor's proposal is between members of two distinct sets.
No. He does not specify that. And there is no reason to do so, except
that it can be used to contradict the ridiculous nonsense that there
are as many fractions as prime numbers.y
But he DOES, as he talks about the two SETS of numbers that are
matched up.
One set and its subset. Dedekind: A system S is said to be /infinite/ if
it is similar to a real part of itself. To consider them as two sets
does not change the numbers of elements.
And yes, the size of the set of all fractions is EXACTLY of the same
size as the set of all Prime Numbers, and that size is Aleph_0.
Wrong. Proof: All prime numbers p are fractions p/1 ∈ ℚ, but 1/2 is not prime.
Regards, WM
On 4/7/24 4:32 AM, WM wrote:
Le 06/04/2024 à 22:03, Richard Damon a écrit :
On 4/6/24 3:40 PM, WM wrote:
Le 06/04/2024 à 15:58, Richard Damon a écrit :
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping, the >>>>>> O's would also remain. All O's! It is th lossless exchange which
proves it.
Cantor's proposal is between members of two distinct sets.
No. He does not specify that. And there is no reason to do so, except
that it can be used to contradict the ridiculous nonsense that there
are as many fractions as prime numbers.y
But he DOES, as he talks about the two SETS of numbers that are
matched up.
One set and its subset. Dedekind: A system S is said to be /infinite/ if
it is similar to a real part of itself. To consider them as two sets
does not change the numbers of elements.
But does affect your logic of pairing.
So, With infinite sets, a proper subset CAN be the same size as its parent.
You are just PROVING you don't understand how infinity works,
Le 06/04/2024 à 17:49, Jim Burns a écrit :
On 4/6/2024 9:44 AM, WM wrote:
But we can use the ordinal axis
as Cantor has described it
0, 1, 2, 3, ...,
ω, ω + 1, ..., ω + k, ...,
ω + ω (= ω2), ω2 + 1, ..
and multiply 0, 1, 2, 3, ..., ω, by 2.
Two times anything on the first row
is something on the first row.
That is contradicted by
That is contradicted by this argument:
Two times all numbers of 1, 2, 3, .., ω
will double the distance between ℕ and ω
to that between 2ℕ and 2ω.
On 4/6/2024 3:49 PM, WM wrote:
Le 06/04/2024 à 17:49, Jim Burns a écrit :
On 4/6/2024 9:44 AM, WM wrote:
But we can use the ordinal axis
as Cantor has described it
0, 1, 2, 3, ...,
ω, ω + 1, ..., ω + k, ...,
ω + ω (= ω2), ω2 + 1, ..
and multiply 0, 1, 2, 3, ..., ω, by 2.
Two times anything on the first row
is something on the first row.
That is contradicted by
...your anti.arithmetism.
The successor operation is closed in
the natural numbers.
Natural.number.addition is closed in
the natural numbers.
Natural.number.multiplication is closed in
the natural numbers.
Those claims are theorems.
In order to be theorems,
it is important to know the meaning of
successor, natural number, addition and multiplication.
If you want to see the proofs, ask.
Two times all numbers of 1, 2, 3, .., ω
will double the distance between ℕ and ω
to that between 2ℕ and 2ω.
No, it won't.
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
Visibleᵂᴹ or darkᵂᴹ,
the successor operation is closed in
the natural numbers.
Multiplication by 2 creates numbers beyond ω
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
WM drivels horrendous bullshit:
Multiplication by 2 creates numbers beyond ω
This statement is totally idiotic, as usual.
there would be numbers immune to multiplication.
Am 09.04.2024 um 14:47 schrieb Tom Bola:
WM drivels horrendous bullshit:
Multiplication by 2 creates numbers beyond ω
This statement is totally idiotic, as usual.
Ach?
(Wer lutscht denn da?)
Moebius schrieb:
Am 09.04.2024 um 14:47 schrieb Tom Bola:
WM drivels horrendous bullshit:
Multiplication by 2 creates numbers beyond ω
This statement is totally idiotic, as usual.
Ach?
(Wer lutscht denn da?)
Ja, ein einzelnes herausgehobenes Statement mit harscher Kritik (und keine ganze Abhandlung wegen der Langeweile so wie du oft) im Rahmen einer Phase, wenn das alle täten, würde WM womöglich die Laune zu posten verlieren...
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
Not correct if
there are all natural numbers such that
no further one exists below ω.
Multiplication by 2
creates numbers beyond ω, or
there would be numbers immune to multiplication.
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
Not correct if there are all natural numbers such that no further one
exists below ω. Multiplication by 2 creates numbers beyond ω, or there would be numbers immune to multiplication.
Regards, WM
On 4/9/2024 3:42 PM, Richard Damon wrote:
Multiplying ANY Natural numbers (which are all finite) by two, results
in another finite Natural Number.
Right! And this new natural number is already in the set of all natural numbers. Why can't WM get this! God damn.
Am 09.04.2024 um 03:54 schrieb Jim Burns:
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k e IN.
(I hope that you don't mind
that slight simplification).
is that I want to express "natural number"k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
On 4/9/2024 8:22 AM, WM wrote:
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
Not correct if
there are all natural numbers such that
no further one exists below ω.
You seem to be saying:
| Not correct if
| not all natural numbers have
| a further one below ω
In other words [1]:
| Not correct if
| ω is finiteⁿᵒᵗᐧᵂᴹ.
Le 09/04/2024 à 20:27, Jim Burns a écrit :
On 4/9/2024 8:22 AM, WM wrote:
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
Not correct if
there are all natural numbers such that
no further one exists below ω.
You seem to be saying:
| Not correct if
| not all natural numbers have
| a further one below ω
All further numbers are multiplied too.
In other words [1]:
| Not correct if
| ω is finiteⁿᵒᵗᐧᵂᴹ.
Not correct if
there is no free space between ℕ and ω.
On 04/10/2024 01:13 PM, Jim Burns wrote:
ℕ = ⟦0,ω⦆
ω is after Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
Avogadroᴬᵛᵒᵍᵃᵈʳᵒ = 6.02214076E23⁶ᐧ⁰²²¹⁴⁰⁷⁶ᴱ²³
and after any other number such that
it and all non.0 numbers.before.it
have predecessors.
Infiniteⁿᵒᵗᐧᵂᴹ is different.
You know, some people have that
Avogadro's number, is sort of a _running_, constant.
What I mean by that is that NIST CODATA every few
years arrives at the current values according to
the energy and configuration of experiment, and,
some of the constants are _running_ in the energy
and configuration of experiment, it results they
get not only more precise, even actually, _smaller_.
Or, you know, larger.
On 4/9/2024 3:47 PM, Chris M. Thomasson wrote:
On 4/9/2024 5:22 AM, WM wrote:
Le 09/04/2024 à 01:54, Jim Burns a écrit :
On 4/8/2024 9:55 AM, WM wrote:
Le 07/04/2024 à 21:47, Jim Burns a écrit :
The successor operation is closed in
the natural numbers.
For visible numbers only.
Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
Not correct if there are all natural numbers such that no further one
exists below ω. Multiplication by 2 creates numbers beyond ω, or
there would be numbers immune to multiplication.
moron.
Are you the never ending story?
https://youtu.be/x1afn71-0sI
Le 30/03/2024 à 23:53, Moebius a écrit :
Hint: The number of elements in E is aleph_0. And 2 * aleph_0 =
aleph_0 + aleph_0 = aleph_0.
That is nonsense, because one number more than ℕ is one number more, independent of what alephs can count.
But it is impossible to have one natural number more than all of ℕ.
Regards, WM
WM was thinking very hard :
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn
be added?
Yes, it is the set of natural numbers.
Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann
lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
natürliche Zahlen als ℕ enthält.
I don't read German.
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn be added?
Am 24.03.2024 um 21:00 schrieb WM:
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none can be >> added?
Yes
Proof:
Def.: n is a /natural number/ iff n e IN.
Hence for any n: If n is a natural number it's in IN. With other words,
all natural numbers are in IN. (Hence no natural number "can be added",
since _all_ natural numbers are already in IN.) qed
Am 24.03.2024 um 22:03 schrieb FromTheRafters:
WM was thinking very hard :
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none cn
be added?
Yes, it is the set of natural numbers.
Indeed!
Wären in der Menge {2, 4, 6, ...} genau so viele Zahlen wie in ℕ, dann >>> lieferte die Vervollständigung {1, 2, 3, 4, 5, 6, ...} "mehr Realität
wie" ℕ = {1, 2, 3, ...}. Dann gäbe es also in der Realität mehr
natürliche Zahlen als ℕ enthält.
I don't read German.
"If there were exactly as many numbers in the set {2, 4, 6, ...} as in
ℕ, then the completion {1, 2, 3, 4, 5, 6, ...} would yield "more reality than" ℕ = {1, 2, 3, ...}. Then there would be more natural numbers in reality than ℕ contains."
Sounds like Chinese to me.
WM was thinking very hard :
Le 02/06/2024 à 05:42, Moebius a écrit :
Am 24.03.2024 um 21:00 schrieb WM:
Does ℕ = {1, 2, 3, ...} contain all natural numbers such that none
can be added?
Yes
Proof:
Def.: x is a /natural number/ iff x e IN.
Hence for any x: If x is a natural number it's in IN. With other
words, all natural numbers are in IN. (Hence no natural number "can
be added", since _all_ natural numbers are already in IN.) qed
If you throw away half of them, namely all odd numbers, then you have
less natural numbers: E = |ℕ|/2. If you add another set of |ℕ|/2 even
numbers, these even numbers cannot be natural numbers. (Because all
even natural numbers are in E.)
什么？
All naturals does not imply finite! Are you a killer whale?
https://youtu.be/NZHHHIMd1TM
| Sysop: | Keyop |
|---|---|
| Location: | Huddersfield, West Yorkshire, UK |
| Users: | 716 |
| Nodes: | 16 (2 / 14) |
| Uptime: | 54:31:43 |
| Calls: | 12,117 |
| Calls today: | 8 |
| Files: | 15,010 |
| Messages: | 6,518,632 |
| Posted today: | 2 |